Calculate the wavelength of radiation emitted by a specific
electron transition in the hydrogen atom.

Emission Spectra. Elements in the gaseous
state that are excited by an electrical current give off electromagnetic
radiation in a process called emission. By passing emitted light through
a prism, an emission spectrum consisting of the wavelengths of emitted electromagnetic
radiation is produced. A continuous spectrum contains
all the wavelengths between the longest and shortest emitted by that substance.
"White" light results from the emission of all wavelengths in the visible part
of the spectrum.

The light emitted from excited atoms in the gaseous state produces a line
spectrum rather than continuous spectrum. A line spectrum consists of several
discrete wavelengths. The observed wavelengths are characteristic of the element
and can serve to identify the presence of the element in a sample.

Emission Spectrum of Hydrogen. A
portion ofthe line spectrum of hydrogen falls in the visible
region, and is called the Balmer series. Figure
7.6(b) of the text shows that the prominent visible lines are at 656 nm,
486 nm, and 434 nm. In all, five spectral series are known for hydrogen. The
Lyman series is in the ultraviolet region, and the Paschen, Brackett, and Pfund
series are in the infrared region

Bohr's Model. In
1913 Niels Bohr produced his famous model of the hydrogen atom. The great achievement
of this model was that it explained the source and the observed wavelengths
of lines in the spectrum of the H atom. Bohr postulated a "solar system" model
for the H atom in which the electron traveled in circular orbits about the proton.
The basic assumptions of Bohr's model were:

The electron moves in a circular orbit about the nucleus. Of the infinite
number of possible orbits only certain orbits with distinct radii are allowed.
As long as the electron remains in an orbit, its energy remains constant.

The energy of the electron increases the farther its orbit lies from the
nucleus. When energy is absorbed by the atom, the electron must jump to a
higher energy orbit. When an electron transition occurs from a higher to a
lower energy orbit, radiation is emitted by the atom.

Bohr was able to calculate the radii of the allowed orbits and their energies.
The energies of the H atom are given by:

where RH is the Rydberg constant, which in units of joules has the
value 2.18 x 10–18 J. The minus sign in the equation simply means
that the energy of the hydrogen atom is lower than that of a completely separated
proton and electron for which the force of attraction is zero. It does not signify
negative energy. The quantity n is called the principal quantum number, and
is an integer; n = 1, 2, 3,…. Inserting values for n into the energy equation
gives the energies of all allowed orbits shown in Figure
7.11 of the text.

The radius of each electron orbit in the hydrogen atom is proportional to
n2. As n increases, the orbit radius increases rapidly. The farther
the electron is from the nucleus, the higher its energy.

Bohr explained that the emission of light, when an electrical current passes
through a gas, is due to the formation of excited atoms. He proposed that in
the emission process an electron drops from a higher to a lower orbit. During
this transition the atom emits radiation. For energy to be conserved, the energy
lost as radiation must equal the difference in energy between the initial energy
level Ei and the final energy level Ef. For the emission
process, ni represents the higher energy level and nf
lower level. Then the change in energy of the atom E
is

E = Ef – Ei

where, according to Bohr's equation, the energy change for the atom is

Bohr's stroke of genius was to equate Eatom
to the energy of a photon of emitted light.

Ephoton = Eatom

Since,

Ephoton

=

hn

=

hc

l

Then, hn = Eatom

and the frequency of light emitted should be given by

n

=

Eatom

h

The wavelengths of emitted light can be calculated from

l

=

c

=

hc

n

Eatom

With this model of the hydrogen atom, Bohr was able to predict the wavelengths
of all the literally hundreds of observed lines in the spectrum of hydrogen!

However, it should be noted that the Bohr model of the atom is not an accurate
representation. The field of quantum mechanics has provided us with tools to
describe the atom, but these are harder to picture.

EXAMPLE The Hydrogen Atom

a. What amount of energy in joules is lost by a hydrogen atom when an electron
transition from n = 3 to n = 2 occurs in the hydrogen atom?E =
x 10^
J

Correct!

Click a Hint button for help.

The energy of the atom is quantized and depends on the orbit of the electron.
Each orbit is assigned a principal quantum number n:

For this transition ni = 3 and nf = 2:

Calculation for (a)

Substituting for RH gives:

E = –0.545 x 10–18 J + 0.242 x 10–18
J

E = –0.303 x 10–18 J = –3.03 x 10–19
J

b. What is the wavelength of the light emitted when the electron transition
n = 3 n = 2 occurs?l =
nm

Correct!

Click a Hint button for help.

The energy lost by the atom appears as a photon of radiation with its respective
frequency and wavelength. Since the energy of the photon is positive, we will
drop the negative sign from the calculation of the wavelength.

Eatom

=

Ephoton

=

hc

l

l

=

hc

Eatom

Calculation for (b)

l

=

hc

=

hc

Eatom

3.03
x 10–19 J

l

=

(6.63
x 10–34 J·s)(3.00 x 108ms–1)

3.03
x 10–19 J

l = 656 x 10–9 m
= 656 nm

OBJECTIVE CHECK

Complete the following questions to check your understanding
of the material. Select the check button to see if you answered correctly.

Which transition for the H atom produces the emission line with the
longest wavelength?

What wavelength of radiation will be emitted when an electron in a hydrogen
atom jumps from the n = 5 to the n = 1 principal energy level? Name the
region of the electromagnetic spectrum corresponding to this wavelength.

A hydrogen emission line in the ultraviolet region of the spectrum at
95.2 nm corresponds to a transition from a higher energy level n to the
n = 1 level. What is the value of n for the higher energy level?

The second line in the Balmer series occurs at 486.1 nm. What is the
energy difference between the initial and final energy levels involved in
the electron transition?